The second term we can integrate by the standard result above. For the first term, we substitute \(u=\tan x\) and get \[
\int \sec^2 x\tan^2 x\:dx = \int u^2\:du = \frac{\tan^3 x}{3}+C.
\] Thus we have \[\int_0^{\pi/4} \sec^4 x \:dx = \left[\frac{\tan^3 x}{3}+\tan x\right]_0^{\pi/4} = \frac{4}{3}.
\]